QJEEM22Ph5977: JEE Main 2022 Physics - Class 12 JEEM22Ph5977 Solution: In the given circuit 'a' is an arbitrary constant. The value of …

${R_{net}} = {{ma} \over 3} + {a \over {2m}}$

$= a\left[ {{m \over 3} + {1 \over {2m}} - {2 \over {\sqrt 6 }} + {2 \over {\sqrt 6 }}} \right]$

$= a\left[ {{{\left( {\sqrt {{m \over 3}} - {1 \over {\sqrt {2m} }}} \right)}^2} + \sqrt {{2 \over 3}} } \right]$

This will be minimum when

$\sqrt {{m \over 3}} = {1 \over {\sqrt {2m} }}$

or $m = \sqrt {{3 \over 2}}$

so $x = 3$